# Picard-Lindelöf Theorem for Stochastic Differential Equations proof

I am self-studying https://sayanmuk.github.io/StochasticAnalysisManifolds.pdf, the following proof can be found on the 9th page of the monograph I just provided the reference. It follows from a previous question that I asked:Intuition behind $Q_t=\sum \langle M^{\alpha},M^{\alpha}\rangle_t+\sum |A^{\alpha}|^3_t+|A^{\alpha}|_t+t$

I am having a problem with two steps of the following theorem proof:

$$\textbf{Theorem }$$ Suppose that $$\sigma$$ is globally Lipschitz and $$X_0$$ is square integrable. Then the stochastic differential equation (SDE) $$(\sigma, Z, X_0)$$ has a unique solution $$X = \{X_t, t \geq 0\}$$.

Proof: In this proof, we use the notation

$$|Y|_{\infty,t} = \max_{0 \leqslant s \leqslant t} |Y_s|$$

for a vector-valued process $$Y$$. We solve the equation by the usual Picard's iteration. Define a sequence $$\{X^n\}$$ of semimartingales by $$X^0_{t} = X_0$$ and

$$$$X^n_{t} = X^0_{t} + \int_{0}^{t} \sigma(X^{n-1}_s) \, dZ_s \tag{1.1.5}$$$$

We claim that $$\{X_n\}$$ converges to a continuous semimartingale $$X$$ which satisfies the equation. Define the increasing process $$Q$$ as in (1.1.3) and let $$\eta$$ be its inverse as before. Each $$\eta_T$$ is a stopping time for fixed $$T$$. By (1.1.4) and the assumption that $$\sigma$$ is globally Lipschitz, we have

$$$$\mathbb{E} \left[ \sup_{0 \leq s \leq \eta_T} \left|X^n - X^{n-1}\right|^2 \right] \leq C_1 \mathbb{E} \left[ \int_0^{\eta_T} \left| \sigma(X^{n-1,s}) \right|^2 \, ds \right]$$$$

\begin{aligned} \leq C_2 \mathbb{E} \left[ \int_{0}^{\eta_T} \left|X^{n-1,s} - X^{n-2,s}\right|^2 dQ_{\eta_T} \right]. \end{aligned}

Making the change of variable $$s = \eta(u)$$ in the last integral and using $$Q_{\eta(u)} = u$$, we have $$$$E \left[ \sup_{0 \leq s \leq \eta_T} \left|X^n - X^{n-1}\right|^2 \right] \leq C_2 E \left[ \int_{0}^{\eta_T} \sup_{0 \leq s \leq \eta_u} \left|X^{n-1}_s - X^{n-2}_s\right|^2 \, du \right].$$$$

For the initial step, from $$$$X^1_{t} - X^0_{t} = \int_{0}^{t} \sigma(X^{0}_s) \, dZ_s = \sigma(Z_0)Z_t.$$$$

and (1.1.4) again we have $$$$E \left[ \sup_{0 \leq s \leq \eta_T} \left|X^{1} - X^{0}\right|^2 \right] \leq C_3 E \left[ \left|X_{0}\right|^2 + 1 \right] Q_{\eta_T} = C_3 T E \left[ \left|X_{0}\right|^2 + 1 \right].$$$$

Questions:

1. I wonder if this is correct:$$X^1_{t} - X^0_{t} = \int_{0}^{t} \sigma(X^{0}_s) \, dZ_s = \sigma(Z_0)Z_t$$. As far as I understand this $$X^{0}_s$$, should be a constant, right? Once it is assumed to be equal to $$X_0$$. If that is so the integral above would be $$\sigma(X_0)Z_t$$ instead. So probably, there was a typo. But my problem relies on what comes next:

2)$$E \left[ \sup_{0 \leq s \leq \eta_T} \left|X^{1} - X^{0}\right|^2 \right] \leq C_3 E \left[ \left|X_{0}\right|^2 + 1 \right] Q_{\eta_T}$$. This step is puzzling me. How does the author get $$|X_{0}|^2+1$$? Is he using some Lipschitz property of $$\sigma$$? How does that relate to the previous step, I mentioned in 1). Since he mentions that we just need to apply the inequality $$E \max_{0\leqslant t\leqslant\tau}|\int_{0}^{t}F_s dZ_s|^2\leqslant C E \int_0^{\tau}|F_s|^2dQ_s$$(see Intuition behind $Q_t=\sum \langle M^{\alpha},M^{\alpha}\rangle_t+\sum |A^{\alpha}|^3_t+|A^{\alpha}|_t+t$ ).

However, if I do apply that inequality I end up with an integral inside expectation of this kind $$\int_0^T |\sigma(X_0)|^2 dQ_{\nu_T}$$. How does this latter ends up being $$(|X|^2+1)Q_{\eta_T}$$

I would appreciate it if someone could clarify those last steps.

1. I wonder if this is correct:$$X^1_{t} - X^0_{t} = \int_{0}^{t} \sigma(X^{0}_s) \, dZ_s = \sigma(Z_0)Z_t$$. As far as I understand this $$X^{0}_s$$, should be a constant, right? Once it is assumed to be equal to $$X_0$$. If that is so the integral above would be $$\sigma(X_0)Z_t$$ instead. So probably, there was a typo. But my problem relies on what comes next:

Yes, but he defined $$X_{s}^{0}=X_{0}$$, so you are both right.

2)$$E \left[ \sup_{0 \leq s \leq \eta_T} \left|X^{1} - X^{0}\right|^2 \right] \leq C_3 E \left[ \left|X_{0}\right|^2 + 1 \right] Q_{\eta_T}$$. This step is puzzling me. How does the author get $$|X_{0}|^2+1$$? Is he using some Lipschitz property of $$\sigma$$? How does that relate to the previous step, I mentioned in 1). Since he mentions that we just need to apply the inequality $$E \max_{0\leqslant t\leqslant\tau}|\int_{0}^{t}F_s dZ_s|^2\leqslant C E \int_0^{\tau}|F_s|^2dQ_s$$(see Intuition behind $Q_t=\sum \langle M^{\alpha},M^{\alpha}\rangle_t+\sum |A^{\alpha}|^3_t+|A^{\alpha}|_t+t$ ). However, if I do apply that inequality I end up with an integral inside expectation of this kind $$\int_0^T |\sigma(X_0)|^2 dQ_{\nu_T}$$. How does this latter ends up being $$(|X|^2+1)Q_{\eta_T}$$

So indeed they apply the 1.1.4 inequality to bound by

$$\int_0^{\eta_T} |\sigma(X_0)|^2 dQ_{t}$$

and yes they use Lipschitz

$$|\sigma(X_0)|^2 \leq |\sigma(X_0)-\sigma(0)|^2 +|\sigma(0)|^2\leq c_{1} |X_{0}^{2}|+c_{2}\leq c_{3}(|X_{0}^{2}|+1).$$